Problem: Solve for $r$, $ \dfrac{r - 8}{r} = \dfrac{3}{4r} + \dfrac{3}{r} $
Answer: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $r$ $4r$ and $r$ The common denominator is $4r$ To get $4r$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ \dfrac{r - 8}{r} \times \dfrac{4}{4} = \dfrac{4r - 32}{4r} $ The denominator of the second term is already $4r$ , so we don't need to change it. To get $4r$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{3}{r} \times \dfrac{4}{4} = \dfrac{12}{4r} $ This give us: $ \dfrac{4r - 32}{4r} = \dfrac{3}{4r} + \dfrac{12}{4r} $ If we multiply both sides of the equation by $4r$ , we get: $ 4r - 32 = 3 + 12$ $ 4r - 32 = 15$ $ 4r = 47 $ $ r = \dfrac{47}{4}$